What is the general solution for 2(sin(x))^2-cos(x)-1=0 ?

The general solution for 2(sin(x))^2-cos(x)-1=0 is x=pi+2pin,x= pi/3+2pin,x=(5pi)/3+2pin

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2(sin(x))^2-cos(x)-1=0

2 (sin (x))^{2} - cos (x) - 1 = 0

x = π + 2 πn, x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

Solution steps

2 (sin (x))^{2} - cos (x) - 1 = 0

Add

cos (x)

to both sides

2 sin^{2} (x) - 1 = cos (x)

Square both sides

(2 sin^{2} (x) - 1)^{2} = cos^{2} (x)

Subtract

cos^{2} (x)

from both sides

(2 sin^{2} (x) - 1)^{2} - cos^{2} (x) = 0

Factor

(2 sin^{2} (x) - 1)^{2} - cos^{2} (x) : (2 sin^{2} (x) - 1 + cos (x)) (2 sin^{2} (x) - 1 - cos (x))

(2 sin^{2} (x) - 1 + cos (x)) (2 sin^{2} (x) - 1 - cos (x)) = 0

Solving each part separately

2 sin^{2} (x) - 1 + cos (x) = 0 or 2 sin^{2} (x) - 1 - cos (x) = 0

2 sin^{2} (x) - 1 + cos (x) = 0 : x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn, x = 2 πn

2 sin^{2} (x) - 1 - cos (x) = 0 : x = π + 2 πn, x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

Combine all the solutions

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn, x = 2 πn, x = π + 2 πn, x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

Verify solutions by plugging them into the original equation

x = π + 2 πn, x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

arctan (3 x) + arctan (x) = \frac{π}{4}

4 sin (θ) = 3 sec (θ), 0 \leq θ < 18 0^{\circ}

\frac{sin ( 8 2 ^{\circ} )}{sin ( x )} = 5

\frac{2 cos ^{2} ( x )}{2 ( 1 - sin ( x ) ) - cos ^{2} ( x )} = 0

2 sin (x) = - 3

What is the general solution for 2(sin(x))^2-cos(x)-1=0 ?
The general solution for 2(sin(x))^2-cos(x)-1=0 is x=pi+2pin,x= pi/3+2pin,x=(5pi)/3+2pin